An introduction to ELECTRE Decision Making Method. This is an outranking method where options outranked each other such that the best option remains.Here both concordance and discordance index was used to find the outranking criteria.

1. ELECTRE
Dr. Mrinmoy Majumder
Course Name : Intro to Multi Criteria Decision Making Methods
Lecture No.10 out of 15
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2. ELECTRE
• ELECTRE is a family of multi-criteria decision analysis methods that
originated under the French School of decision making in the mid-
1960s.
• ELECTRE stands for: ELimination Et Choix Traduisant la REalité
(ELimination and Choice Expressing REality).
• The method was invented by Bernard Roy and his colleagues at
SEMA consultancy company.

3. Example of
ELECTRE
Decision Goal : To buy a car
Criteria : Cost and Speed
Alternatives : Mercedes Benz(M),
Jaguar(J), Toyota(T)
Aggregation Methods to be used :
ELECTRE

4. Step 1 : Development of the Alternative
Indicator Matrix
• First step of ELECTRE method is to create a Alternative-Indicator
Matrix :
Indicator
Cost (in
Lakh Rs.)
Speed (in km/hr)
Alternative
Mercedes Benz 80 200
Jaguar 100 300
Toyota 120 250

5. Step 2 : Development of the Normalized Indicator
Matrix or Normalized Decision Matrix
Matrix from Step 1 Indicator
Cost (in
Lakh Rs.)
Speed (in
km/hr)
Alternative
Merced
es Benz
80 200
Jaguar 100 300
Toyota 120 250
Square each value of the
indicators and add column
wise. Then find the square
root of the summation.
Divide each value of the
Indicators with the square
root.
Indicator
Cost (in
Lakh Rs.)
Speed (in
km/hr)
Alternative
Mercedes
Benz 6400 40000
Jaguar 10000 90000
Toyota 14400 62500
Column wise Sum 30800 192500
Square Root of the Sum 175.499 438.748

6. Step 2 : Contd.
• The Normalized Decision Matrix
Indicator
Cost Speed
Alternative
Mercedes
Benz
0.456 0.456
Jaguar 0.570 0.684
Toyota 0.684 0.570

7. Step 3 : Development of the Weighted
Normalized Decision Matrix
Indicator
Cost
(Weight
of
Indicator :
0.600)
Speed
(Weight
of
Indicator :
0.400)
Alternative
Mercedes
Benz
0.456 0.456
Jaguar 0.570 0.684
Toyota 0.684 0.570
Multiply the weight of
indicator of each
column with each value
of the alternatives for
that indicator to find
the weighted value of
the indicators for the
alternatives
Indicator
Cost
(Weight of
Indicator :
0.600)
Speed
(Weight of
Indicator :
0.400)
Alternative
Mercedes
Benz
0.274 0.182
Jaguar 0.342 0.274
Toyota 0.410 0.228

8. Step 4 : Development of the Concordance
Matrix Each alternative is compared with the
other alternative with respect to its
normalized value for the indicators.
If normalized value of M and J is
compared with respect to Cost indicator
then M < J, thus 0 is written. M is less
than J for Speed indicator as well. Thus
the value in the matrix will be 0.However
when J is compared with M, J>M for both
Cost and Speed Indicator. So the weight
of both the indicator will be added and
shown in that cell of the matrix.
Mercedes
Benz(M) Jaguar(J)
Toyota(T)
Mercedes Benz(M) 0 0 0
Jaguar(J) =0.6+0.4 0 =0+0.4
Toyota(T) =0.6+0.4 =0.6+0 0
Matrix from Step 3
Indicator
Cost
(Weight of
Indicator :
0.600)
Speed
(Weight of
Indicator :
0.400)
Alternative
Mercedes
Benz(M)
0.274 0.182
Jaguar(J) 0.342 0.274
Toyota(T) 0.410 0.228

9. Step 5 : Concordance Matrix
Mercedes
Benz Jaguar
Toyota
Mercedes Benz 0 0 0
Jaguar 1 0 0.4
Toyota 1 0.6 0
Column wise Sum
= 0+1+1 =
2
= 0+0+0.6 = 0.6 =0+0.4+0 = 0.4
Total : =2 + 0.6 + 0.4 = 3
Total/Number of Values in the Matrix = 3/4 = 0.75
Matrix from Step 4 Mercedes
Benz Jaguar
Toyota
Mercedes Benz 0 0 0
Jaguar =0.6+0.4 0 =0+0.4
Toyota 0.6+0.4 =0.6+0 0
1 2
3 4
Only the cell which depicts the
comparison between J with M,T with M,T
with J and J with T has real values. As a
result number of values in the matrix is 4

10. Step 5 : Contd.
Concordance Set : If C bar
(see last row of matrix 4) is less
than the value in the cell of the
matrix then the value will be
replaced by 1 otherwise if R is
greater than the real value in
the cell then 0 is used instead
of the existing value.
Mercedes
Benz Jaguar
Toyota
Mercedes Benz 0 0 0
Jaguar 1 0 0
Toyota 1 0 0
Matrix 4 Mercedes
Benz Jaguar
Toyota
Mercedes Benz 0 0 0
Jaguar 1 0 0.4
Toyota 1 0.6 0
Column wise
Sum
2 0.6 0.4
Total : =2 + 0.6 + 0.4 = 3
Total / (Number
of cells in the
Matrix where a
real number
exist) = C bar
= 3/4 = 0.75

11. Step 6 : Development of the Discordance
Matrix
Matrix from Step 3
Indicator
Cost
(Weight of
Indicator :
0.600)
Speed
(Weight of
Indicator :
0.400)
Alternative
Mercedes
Benz(M)
0.274 0.182
Jaguar(J) 0.342 0.274
Toyota(T) 0.410 0.228
The normalized value of each alternative
for each indicator is deducted from the
values of other alternatives for the same
indicator
Cost
Speed
M-J = 0.274 - 0.342 = 0.182 - 0.274
M-T = 0.274 - 0.410 = 0.182 - 0.228
J-M = 0.342 - 0.274 = 0.274 - 0.182
J-T = 0.342 - 0.410 = 0.274 - 0.228
T-M = 0.410 – 0.274 = 0.228 - 0.182
T-J = 0.410 – 0.342 = 0.228 - 0.274

12. Column : 1
The normalized value of each
alternative for each indicator is
deducted from the values of other
alternatives for the same indicator
Column : 2
Cost
Column : 3
Speed
Column : 4
Find the
maximum
value in the
row
(A)
Column : 5
Find the
maximum
negative
value or if
there is no
negative,
then use the
maximum
value of the
row(B)
Column :
6
(B)÷(A)
M-J -0.068 -0.091 0.091 0.091 1
M-T -0.137 -0.046 0.137 0.137 1
J-M 0.068 0.091 0.091 0.091 1
J-T -0.068 0.046 0.068 0.068 1
T-M 0.137 0.046 0.137 0.137 1
T-J 0.068 -0.046 0.068 0.046 0.667
Rough Set Matrix

13. Discordance Set : If D bar
(see last row of matrix 5) is less
than the value in the cell of the
matrix then the value will be
replaced by 1 otherwise if R is
greater than the real value in
the cell then 0 is used instead
of the existing value.
Mercedes
Benz Jaguar
Toyota
Mercedes Benz 0 1 1
Jaguar 1 0 1
Toyota 1 0 0
Matrix 5 : Matrix
from Step 4 can be
rewritten by using
the values from
Column 6 of Rough
Set Matrix
Mercedes
Benz(M) Jaguar(J)
Toyota(T)
Mercedes Benz(M) 0 1 1
Jaguar(J) 1 0 1
Toyota(T) 1 0.667 0
Column wise Sum 2 1.667 2
Total : =2 + 1.667 + 2 = 5.667
Total / (Number of
cells in the Matrix
where a real
number exist) = D
bar
= 5.667/6 = 0.945

14. Concordance
Set
(C)
Mercedes
Benz Jaguar
Toyota
Mercedes Benz 0 0 0
Jaguar 1 0 0
Toyota 1 0 0
C (AND or × ) D Mercedes
Benz
Jaguar Toyota
Mercedes Benz 0 0 0
Jaguar 1 0 0
Toyota 1 0 0
Discordance Set(D) Mercedes
Benz Jaguar
Toyota
Mercedes Benz 0 1 1
Jaguar 1 0 1
Toyota 1 0 0
0 0AND
or ×
= 0
1 1AND
or ×
= 1
AND
or ×
EXAMPLE
It Implies that :
J > M and T> M
Or
J and T > M
0 1AND
or ×
= 0

15. Thank you